Difference between revisions of "2010 AMC 10B Problems/Problem 19"

Revision as of 17:32, 4 May 2024

Problem

A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$ ?

$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution 1

The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$

Because $OA=4\sqrt{3}=\sqrt{48} < \sqrt{156}, A$ must be in the interior of circle $O.$

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; real r=sqrt(156); pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147)); pair O=(0,0); pair X=(0,7sqrt(3)); path outer=Circle(O,r); draw(outer); draw(A--B--C--cycle); draw(O--X); draw(O--B); pair[] ps={A,B,C,O,X}; dot(ps); label("$A$",A,SE); label("$B$",B,NW); label("$C$",C,NE); label("$O$",O,S); label("$X$",X,N); label("$s$",A--C,SE); label("$\frac{s}{2}$",B--X,N); label("$\frac{s\sqrt{3}}{2}$",A--X,NE); label("$\sqrt{156}$",O--B,SW); label("$4\sqrt{3}$",A--O,E); [/asy]$

Let $s$ be the side length of the triangle, the unknown value, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equilateral, $BX=\frac{s}{2}$ and $AX=\frac{s\sqrt{3}}{2}.$ We are given $AO=4\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$

$\begin{align*} (\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\ 156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\ 0 &= s^2 + 12s - 108\\ 0 &= (s-6)(s+18)\\ s &= \boxed{\textbf{(B)}\ 6} \end{align*}$

Solution 2

We can use the same diagram as Solution 1 and label the side length of $\triangle ABC$ as $s$ . Using congruent triangles, namely the two triangles $\triangle BOA$ and $\triangle COA$ , we get that $\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150$ . From this, we can use the Law of Cosines, to get $s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2$ Simplifying, we get $s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0$ We can factor this to get $(x-6)(x+18)$ Lengths must be non-negative, so the answer is $\boxed{\textbf{(B)}\ 6}$ ~bryan gao

Video Solution

https://youtu.be/FQO-0E2zUVI?t=906

~IceMatrix

@@ Line 39: / Line 39: @@
 </asy></center>
-Let <math>s</math> be the unknown value, the sidelength of the triangle, and let <math>X</math> be the point on <math>BC</math> where <math>OX \perp BC.</math> Since <math>\triangle ABC</math> is equilateral, <math>BX=\frac{s}{2}</math> and <math>AX=\frac{s\sqrt{3}}{2}.</math> We are given <math>AO=4\sqrt{3}.</math> Use the [[Pythagorean Theorem]] and solve for <math>s.</math>
+Let <math>s</math> be the side length of the triangle, the unknown value, and let <math>X</math> be the point on <math>BC</math> where <math>OX \perp BC.</math> Since <math>\triangle ABC</math> is equilateral, <math>BX=\frac{s}{2}</math> and <math>AX=\frac{s\sqrt{3}}{2}.</math> We are given <math>AO=4\sqrt{3}.</math> Use the [[Pythagorean Theorem]] and solve for <math>s.</math>
 <cmath>\begin{align*}

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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